\begin{align}
P_{cec} = \left\{\vc{x} \in \mathbb{R}^{\left|E\right|} : \sum_{e \in E} x_e = n-1, \quad \sum_{e \in C} x_e \leq \left|C\right|-1, \forall C \subseteq E, \left|C\right| \geq 2, C \; \mbox{forms a cycle}, \quad \vc{0} \leq \vc{x} \leq \vc{1} \right\} \label{Pcec} \\ 
P_{sub} =  \left\{\vc{x} \in \mathbb{R}^{\left|E\right|} : \sum_{e \in E} x_e = n-1, \quad \sum_{e \in E\left(S\right)} x_e \leq \left|S\right|-1, \forall S \subseteq V, S \neq \varnothing, \quad \vc{0} \leq \vc{x} \leq \vc{1}\right\} \label{Psub}
\end{align}

We disprove the relationship $P_{\textrm{cec}} \subseteq P_{\textrm{sub}}$ by giving the counterexample in Fig. \ref{fig:ex9graph}.

\begin{figure}[ht!]
\centering
\includegraphics[height=40mm]{ex9graph.png}
\caption{Counterexample that proves $P_{\textrm{cec}} \nsubseteq P_{\textrm{sub}}.$}
\label{fig:ex9graph}
\end{figure}

The assignment for $\vc{x}$ fulfills $\vc{x} \in P_{\textrm{cec}}$ and $\vc{x} \notin P_{\textrm{sub}}$.

There are five nodes in the graph and $\sum_{e\in E} x_e = 4$. There are three cycles in the graph: two triangles $C_{1,2}$ for which $\sum_{e\in C_{1,2}} x_e \leq 2$ and a square $C_3$ for which $\sum_{e\in C_3} x_e \leq 3$. Therefore $\vc{x} \in P_{\textrm{cec}}$.

But considering the node set $S$ that consists of the four corners of the square, we find that $\sum_{e \in E(S)} x_e = 3.5 \nleq |S|-1=3$. Therefore $\vc{x} \notin P_{\textrm{sub}}$.